How many NESA Chemistry questions cover Chemical Synthesis and Design?

AusGrader has 34 NESA Chemistry questions on Chemical Synthesis and Design, all with instant AI grading and detailed marking feedback.

NESA (HSC) Chemistry — Chemical Synthesis and Design Questions & Answers | AusGrader

Q7

2024

QCAA

Paper 1

1 mark

Q7

1 mark

Which option is a principle of green chemistry?

A

avoid chemical derivatives

B

decrease energy efficiency

C

prevent catalytic reactions

D

minimise atom economy

Q12

2021

QCAA

Paper 1

1 mark

Q12

1 mark

Green chemistry principles include the design of chemical synthesis processes that

A

use renewable raw materials and minimise unwanted products.

B

use renewable raw materials and minimise unwanted reactants.

C

use non-renewable raw materials and minimise unwanted products.

D

use non-renewable raw materials and minimise unwanted reactants.

Q30

2025

NESA

5 marks

Q30

Phosgene is used in industry as a starting material to synthesise useful polymers. Phosgene ( $\text{Cl}_2\text{CO}$ ) is a gas at room temperature and is highly toxic.

Q30a

2 marks

Justify a suitable precaution when using phosgene.

Phosgene is dangerous to the environment because it is toxic. Therefore, it is important to ensure it does not leak from the reaction vessel, which requires constant monitoring.

Marking Criteria

Descriptor	Marks
Justifies a named precaution	2
Provides any relevant information	1
None of the above	0

Q30b

3 marks

Phosgene is synthesised by the reaction of carbon monoxide (CO) and chlorine ( $\text{Cl}_2$ ) in the gas phase.

\text{Cl}_2(g) + \text{CO}(g) \rightleftharpoons \text{Cl}_2\text{CO}(g)

Explain why an excess of carbon monoxide and a catalyst are used in the industrial synthesis of phosgene.

The use of a catalyst saves time and money because it increases the rate of reaction by lowering the activation energy. The use of a large excess of CO causes the equilibrium to shift to the products side, therefore improving the yield of phosgene.

Marking Criteria

Descriptor	Marks
Explains the use of a catalyst AND Explains the use of a large excess of CO	3
Describes how BOTH features affect the reaction but does NOT link cause and effect OR Explains how one feature affects the reaction	2
Provides any relevant information	1
None of the above	0

Q3

2020

QCAA

Paper 2

9 marks

Q3

Ethanol can be produced by the fermentation of glucose or the hydration of ethene.

Q3a

3 marks

Describe the production of ethanol by fermentation of glucose by writing a balanced equation and indicating if a catalyst is required.

C $_6$ H $_{12}$ O $_6$ (aq) $\xrightarrow{\text{yeast}}$ 2CH $_3$ CH $_2$ OH(aq) + 2CO $_2$ (g)

Marking Criteria

Descriptor	Marks
provides correct reactants and products	1
correctly balances the equation	1
indicates that yeast is required as a catalyst	1

Q3b

2 marks

Calculate the atom economy for the production of ethanol by fermentation of glucose.

Molar mass (ethanol) = 46.08 g
Molar mass (glucose) = 180.18 g
atom economy = $\frac{2 \times 46.08}{180.18} \times 100$

atom economy = $\frac{2 \times 46.08}{180.18} \times 100 = 51.148 \% \approx 51\%$
Atom economy = 51%

Marking Criteria

Descriptor	Marks
shows substitution correctly performed	1
determines atom economy	1

Q3c

2 marks

In terms of atom economy, determine which process for the production of ethanol (i.e. hydration of ethene or fermentation of glucose) is greener.

Hydration atom economy = 100%
Fermentation atom economy = 51%
Therefore, production of ethene by hydration is greener.

Marking Criteria

Descriptor	Marks
determines atom economy for hydration reaction is 100%	1
identifies that hydration reaction is greener	1

Q3d

2 marks

Identify two principles of green chemistry, other than atom economy, that make the production of ethanol by fermentation greener than by hydration.

Descriptor	Marks
provides use of renewable feedstocks	1
provides design for energy efficiency	1

Q38

2021

SCSA

12 marks

Q38

12 marks

Sulfuric acid is manufactured by the Contact process, the steps of which are outlined below.

Step One: Molten sulfur is burned in air at approximately 1000 °C:

S(l) + O_2(g) \rightarrow SO_2(g) + 297 kJ

Step Two: The resulting sulfur dioxide is converted to sulfur trioxide as shown in the following equilibrium reaction. It is conducted at a temperature of about 450 °C with a $V_2O_5$ catalyst at a pressure of between 100 and 200 kPa:

2 SO_2(g) + O_2(g) \rightleftharpoons 2 SO_3(g) + 198 kJ

Step Three: The resulting sulfur trioxide is absorbed into sulfuric acid, producing oleum ( $H_2S_2O_7$ ). Water is added to the oleum, producing 18 mol L $^{-1}$ sulfuric acid:

SO_3(g) + H_2SO_4(l) \rightarrow H_2S_2O_7(l)

H_2S_2O_7(l) + H_2O(l) \rightarrow 2 H_2SO_4(aq)

Use your understanding of collision theory and chemical equilibrium to discuss the reaction conditions for Steps 1 and 2 of the Contact process, given that the aim is to produce the greatest yield in the shortest time. In your discussion, also address economic concerns where appropriate.

High temperature increases the average kinetic energy of the particles, which means that the particles collide more frequently. Also, more of these collisions will have energy higher than the activation energy, which means a greater proportion of collisions are successful, and the reaction rate increases. The vanadium catalyst increases the rate of the forward reaction, and also the rate of the reverse reaction to an equal extent, as it provides an alternative pathway with a lower activation energy. Therefore, a greater proportion of the particles will have sufficient energy to react when they collide. High pressure or concentration has more particles per unit volume and so there is a higher frequency of collisions, and the reaction rate increases. As Step 1 is a combustion reaction, it essentially goes to completion at the high temperature and does not require a catalyst or high pressure. For Step 2, high temperature, high pressure and a catalyst would favour a high rate.

For equilibrium, which is only considered for Step 2, high temperature favours the reverse reaction because it is endothermic, and this decreases the SO $_3$ (g) yield, which is not desired. A low temperature decreases the rate of reaction, which is also not desired. A high pressure favours the forward reaction because there are a greater number of moles of gas reactants, increasing the SO $_3$ (g) yield which is desired.

Economically, high pressures are costly and dangerous.

Therefore, for Step 2, a compromise is required between the high temperature for rate and the low temperature for yield. A compromise is also required between the cost of higher pressures and the pressure that allows a satisfactory yield and rate.

Marking Criteria

Rates

Descriptor	Marks
1 mark each for any of the following, up to a maximum of 6 marks: Explains that high temperature increases the average kinetic energy of the particles, which means that the particles collide more frequently; Explains that more of these collisions will have energy higher than the activation energy, which means a greater proportion of collisions are successful, and the reaction rate increases; Describes that the vanadium catalyst increases the rate of the forward reaction (and also the rate of the reverse reaction to an equal extent) as it provides an alternative pathway with a lower activation energy; States that a greater proportion of the particles will have sufficient energy to react when they collide; Explains that high pressure (concentration) has more particles per unit volume and so there is a higher frequency of collisions, and the reaction rate increases; Identifies that as Step 1 is a combustion reaction, it essentially goes to completion at the high temperature (and does not require a catalyst or high pressure); States that for Step 2, high temperature, high pressure and catalyst would favour high rate.	6

Descriptor

Marks

1 mark each for any of the following, up to a maximum of 6 marks: Explains that high temperature increases the average kinetic energy of the particles, which means that the particles collide more frequently; Explains that more of these collisions will have energy higher than the activation energy, which means a greater proportion of collisions are successful, and the reaction rate increases; Describes that the vanadium catalyst increases the rate of the forward reaction (and also the rate of the reverse reaction to an equal extent) as it provides an alternative pathway with a lower activation energy; States that a greater proportion of the particles will have sufficient energy to react when they collide; Explains that high pressure (concentration) has more particles per unit volume and so there is a higher frequency of collisions, and the reaction rate increases; Identifies that as Step 1 is a combustion reaction, it essentially goes to completion at the high temperature (and does not require a catalyst or high pressure); States that for Step 2, high temperature, high pressure and catalyst would favour high rate.

6

Equilibrium

Descriptor	Marks
1 mark each for any of the following (only considered for Step 2), up to a maximum of 3 marks: Explains that high temperature favours the reverse reaction because it is endothermic, and this decreases the SO3(g) yield; States that a low temperature decreases the rate of reaction; Explains that a high pressure favours the forward reaction because there are a greater number of moles of gas reactants, increasing the SO3(g) yield.	3

Descriptor

Marks

1 mark each for any of the following (only considered for Step 2), up to a maximum of 3 marks: Explains that high temperature favours the reverse reaction because it is endothermic, and this decreases the SO3(g) yield; States that a low temperature decreases the rate of reaction; Explains that a high pressure favours the forward reaction because there are a greater number of moles of gas reactants, increasing the SO3(g) yield.

3

Economics

Descriptor	Marks
States that high pressures are costly (and dangerous).	1

Compromise

Descriptor	Marks
1 mark each for any of the following, up to a maximum of 2 marks: Identifies that for Step 2, a compromise is required between the high temperature for rate and the low temperature for yield; Identifies that a compromise is also required between the cost of higher pressures and the pressure that allows a satisfactory yield and rate.	2

NESA Chemistry Chemical Synthesis and Design

Rates

Equilibrium

Economics

Compromise

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